Problem: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 5}{x + 7} = \dfrac{2x + 58}{x + 7}$
Explanation: Multiply both sides by $x + 7$ $ \dfrac{x^2 - 5}{x + 7} (x + 7) = \dfrac{2x + 58}{x + 7} (x + 7)$ $ x^2 - 5 = 2x + 58$ Subtract $2x + 58$ from both sides: $ x^2 - 5 - (2x + 58) = 2x + 58 - (2x + 58)$ $ x^2 - 5 - 2x - 58 = 0$ $ x^2 - 63 - 2x = 0$ Factor the expression: $ (x + 7)(x - 9) = 0$ Therefore $x = -7$ or $x = 9$ However, the original expression is undefined when $x = -7$. Therefore, the only solution is $x = 9$.